JEE Main Alternating Current PYQ

41. (JEE Main 2020 (Online) 2nd September Evening Slot)

An inductance coil has a reactance of 100 $Ω$ . When an AC signal of frequency 1000 Hz is applied to the coil, the applied voltage leads the current by 45^o. The self-inductance of the coil is

A. 6.7 $\times$ 10^–7 H

B. 1.1 $\times$ 10^–1 H

C. 5.5 $\times$ 10^–5 H

D. 1.1 $\times$ 10^–2 H

Correct Option is (D)

L-R circuit :

tan 45^o = $\frac{X_{L}}{R}$

$\Rightarrow$ 1 = $\frac{X_{L}}{R}$

$\Rightarrow$ X_L = R

Now Z = $\sqrt{R^{2} + X_{L}^{2}}$

or Z = $\sqrt{X_{L}^{2} + X_{L}^{2}} = \sqrt{2 X_{L}^{2}}$ = $\sqrt{2} X_{L}$

$\Rightarrow$ 100 = $\sqrt{2} X_{L}$

X_L = $\frac{100}{\sqrt{2}}$

$\Rightarrow$ $ω L$ = $\frac{100}{\sqrt{2}}$

$\Rightarrow$ L = $\frac{100}{\sqrt{2} \times 2 \times 3.14 \times 1000}$

= 1.1 $\times$ 10^–2 H

42. (JEE Main 2020 (Online) 9th January Evening Slot)

In LC circuit the inductance L = 40 mH and
capacitance C = 100 $μ$ F. If a voltage
V(t) = 10sin(314t) is applied to the circuit, the
current in the circuit is given as :

A. 0.52 cos 314 t

B. 5.2 cos 314 t

C. 0.52 sin 314 t

D. 10 cos 314 t

Correct Option is (A)

JEE Main 2020 (Online) 9th January Evening Slot Physics - Alternating Current Question 81 English Explanation

Z = x_C – x_L

= $\frac{1}{ω C} - ω L$

= $\frac{1}{314 \times 100 \times 10^{- 6}} - 314 \times 40 \times 10^{- 3}$

= 19.28 $Ω$

As V_m = I_mZ

$\Rightarrow$ 10 = I_m $\times$ 19.28

$\Rightarrow$ I_m = $\frac{10}{19.28}$ = 0.52 A

$∴$ I = 0.52 sin(314t + $\frac{π}{2}$ )

= 0.52 cos(314t)

43. (JEE Main 2019 (Online) 8th April Evening Slot)

A circuit connected to an ac source of emf e = e₀sin(100t) with t in seconds, gives a phase difference of $π$ /4 between the emf e and current i. Which of the following circuits will exhibit this ?

A. RC circuit with R = 1 k $Ω$ and C = 1μF

B. RL circuit with R = 1k $Ω$ and L = 1mH

C. RC circuit with R = 1k $Ω$ and C = 10 μF

D. RL circuit with R = 1 k $Ω$ and L = 10 mH

Correct Option is (C)

Given phase difference = $\frac{π}{4}$ and $ω$ = 100 rad/s

$\Rightarrow$ Reactance (X) = Resistance (R) Now by checking option.

Option (A)
R = 1000 $Ω$ and X_c = $\frac{1}{10^{- 6} \times 100} = 10^{4} Ω$

Option (B)
R = 10³ $Ω$ and X_L = $10^{- 3} \times 100 = 10^{- 1} Ω$

Option (C)
R = 10³ $Ω$ and X_c = $\frac{1}{10 \times 10^{- 6} \times 100} = 10^{3} Ω$

Option (D)
R = 10³ $Ω$ and X_L = $10 \times 10^{- 3} \times 100 = 1 Ω$

44. (AIEEE 2005)

The phase difference between the alternating current and $e m f$ is $\frac{π}{2} .$ Which of the following cannot be the constituent of the circuit?

A. $R, L$

B. $C$ alone

C. $L$ alone

D. $L, C$

Correct Option is (A)

The phase difference between the alternating current and emf in an AC circuit depends on the components in the circuit:

In a purely resistive ( $R$ ) circuit, the current and emf are in phase, meaning the phase difference is $0$ .

In a purely inductive ( $L$ ) circuit, the current lags behind the emf by $\frac{π}{2}$ , meaning the phase difference is $\frac{π}{2}$ .

In a purely capacitive ( $C$ ) circuit, the current leads the emf by $\frac{π}{2}$ , again meaning the phase difference is $\frac{π}{2}$ .

In an $L$ - $R$ or $L$ - $C$ circuit, the phase difference depends on the relative values of $L$ , $R$ , and $C$ and can be anywhere between $0$ and $\frac{π}{2}$ .

Therefore, if the phase difference between the alternating current and emf is $\frac{π}{2}$ , then the circuit cannot contain only a resistor ( $R$ ) since that would give a phase difference of $0$ . So, the answer is Option A: $R, L$ . The phase difference would not be $\frac{π}{2}$ if the circuit contains both a resistor and an inductor.

45. (AIEEE 2005)

A circuit has a resistance of $12$ $o h m$ and an impedance of $15$ $o h m$ . The power factor of the circuit will be

A. $0.4$

B. 0.8

C. 0.125

D. 1.25

Correct Option is (D)

Power factor $= \cos ϕ = \frac{R}{Z} = \frac{12}{15} = \frac{4}{5} = 0.8$